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```
ABC is a triangle inscribed in a circle, AC being the diameter of the circle. The length of AC is as much more than the length of BC as the length of BC is more than the length of AB. Find AC:AB.```
8 years ago 30 Points
```

Dear Vipul,
Solution:- Given, AC = a BC and BC = a AB. We want to find AC: AB  = a2
Since, AC is the diameter of the circle, it subtends 900 angle at B. Let, AC = 'd' and angle C be 'x'. Then,
AB = d sinx , and BC = d cosx
So, AC = a BC will give 1 = a cosx     or    cosx = 1/a
And, BC = a AB will give cos x = a sinx
Since, cos x = 1/a, sin x = √(a2 -1) / a
So, 1 = a √(a2 -1)          or         a4 - a2 -1 = 0
On solving quad. eq., we'll get - a2 = (√5 +1) / 2  [ANS]
Please feel free to post as many doubts on our discussion forum as you can. If you find any questionDifficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best Vipul!!!Regards,Askiitians ExpertsPriyansh Bajaj

```
8 years ago
```							Dear Priyansh,
I would like to correct your solution as follows:-
As per question we have to assume that if AB = X and BC = X+k then AC = BC+ k or AC = X+2k. Then we have to find AC:BC which is equal to 5:3. I would request you to find it out for my son Vipul.
Thanking you.
G.P.Agrawal
F/o Master Vipul Agrawal.
```
8 years ago
```							ΔABC is an right angle triangle as Ac is the diameter of the circle. Now it is given that AC-BC=BC-ABAC+AB=2BCAC=2BC-AB............(1)AB²+BC²=AC². .........(Pythagoras theorem) AB²+BC²=(2BC-AB)²............(using eqn.  1)AB²+BC²=4BC²+AB²-4*BC*ABBC²-4BC²=-4*BC*AB3BC²=4*BC*AB3BC²/4BC=AB3/4BC=AB........(2)AB²+BC²=AC²(3/4BC)²+BC²=AC².......(Using eqn. 2)9/16BC²+BC²=AC²9/16BC²+16/16BC²=AC²25/16BC²=AC²5/4BC=AC........(3)AC:AB=AC/AB(5/4BC)/(3/4BC)..........(Using eqn.  2 &3)5/3=AC/AB=AC:AB
```
10 months ago
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